`P=[{:(1,1,1),(0,2,2),(0,0,3):}]Q=[{:(2,x,x),(0,4,0),(x,x,6):}]` and `R=PQP^(-1)` then which are correct

Options

A. det R=det`[{:(2,x,x),(0,4,0),(x,x,5):}]+8` for all `x in R`
B. for x=1 thre exists a unit vector `alphahati+betahatj+gammahatk` for which are `R[{:(alpha),(beta),(gamma):}]=[{:(0),(0),(0):}]`
C. for x=0 if `R[{:(1),(a),(b):}]=6[{:(1),(a),(b):}]` then a+b=5
D. There exists a real number x such that `PQ=QP`

Answer

A:C

Explanation

`detR=detPxxdetQxxdet(P^(-1))`
`impliesdetR=detQ=4(12-x^(2))=48-4x^(2)`
Now `[{:(2,x,x),(0,4,0),(x,x,5):}]=4(10-x^(2))=40-4x^(2)`
`impliesdetR=det({:(2,x,x),(0,4,0),(x,x,5):})+8 AA x in R`
`implies ` Option (A) is correct
At x=1, det Q=48-4=44=det R
Because det `R ne 0` so
`R[{:(alpha),(beta),(gamma):}]=[{:(0),(0),(0):}]impliesalpha=beta=gamma=0`
So `alphahati+betahatj+gammahatk` is not a unit vector
Option (B) is wrong
`P=[{:(1,1,1),(0,2,2),(0,0,3):}],Q(x=0)=[{:(2,0,0),(0,4,0),(0,0,6):}]`
`R=PQP^(-1)`
`=[{:(2,4,6),(0,8,12),(0,0,18):}](1)/(6)[{:(6,-3,0),(0,3,-2),(0,0,2):}]=(1)/(6)[{:(12,6,4),(0,24,8),(0,0,36):}]`
`R=[{:(2,1,2//3),(0,4,4//3),(0,0,6):}]`
`(R-6l)((1),(a),(b))=[{:(-4,1,2//3),(0,-2,4//3),(0,0,0):}]((1),(a),(b))`
`-4+a+(2b)/(3)=0`
`-2a+(4)/(3)=0`
`-2+(4)/(3)b=0`
`-8+(8)/(3)=0impliesb=3`
`a=2` (C) correct
`PQ=QPimpliesa_(12)` for both are same
`impliesx+4+x=2+2x+0implies x in phiimplies` no value exist
`implies` option (D) is wrong